Solve the logarithmic equation. Round your answer to two decimal places. log ⁡ 2 x + log ⁡ 2 ( x + 1 ) − log ⁡ 2 3 = 4 \log _ { 2 } x + \log _ { 2 } ( x + 1 ) - \log _ { 2 } 3 = 4 lo g 2 x + lo g 2 ( x + 1 ) − lo g 2 3 = 4 A) x = 2.00 x = 2.00 x = 2.00 B) x = − 3.00 ; x = 2.00 x = - 3.00 ; x = 2.00 x = − 3.00 ; x = 2.00 C) x = 6.45 x = 6.45 x = 6.45 D) x = − 7.45 ; x = 6.45 x = - 7.45 ; x = 6.45 x = − 7.45 ; x = 6.45 E) none of these

Question

Solve the logarithmic equation. Round your answer to two decimal places.   log ⁡ 2 x + log ⁡ 2 ( x + 1 )  − log ⁡ 2 3 = 4 \log _ { 2 } x + \log _ { 2 } ( x + 1 )  - \log _ { 2 } 3 = 4 lo g 2 ​ x + lo g 2 ​ ( x + 1 )  − lo g 2 ​ 3 = 4 A)    x = 2.00 x = 2.00 x = 2.00 B)    x = − 3.00 ; x = 2.00 x = - 3.00 ; x = 2.00 x = − 3.00 ; x = 2.00 C)    x = 6.45 x = 6.45 x = 6.45 D)    x = − 7.45 ; x = 6.45 x = - 7.45 ; x = 6.45 x = − 7.45 ; x = 6.45 E) none of these

Zeenab Noral · Accepted Answer

Final Answer : C, Explanation :   Using logarithm rules, we can simplify the equation as follows: log ⁡ 2 ( x ( x + 1 ) 3 ) = 4 \log_2\left(\frac{x(x+1)}{3}\right)=4 lo g 2 ​ ( 3 x ( x + 1 ) ​ ) = 4 x ( x + 1 ) 3 = 2 4 = 16 \frac{x(x+1)}{3}=2^4=16 3 x ( x + 1 ) ​ = 2 4 = 16 x 2 + x − 48 = 0 x^2+x-48=0 x 2 + x − 48 = 0 Solving for x, we get $x=6.45$ or $x=-7.45$. However, since we cannot take the logarithm of a negative number, we discard the negative solution. Therefore, the only valid solution is $x=6.45$.

Solve the logarithmic equation. Round your answer to two decimal places. $log _ { 2 } x + \log _ { 2 } ( x + 1 ) - \log _ { 2 } 3 = 4$

A) $x = 2.00$
B) $x = - 3.00; x = 2.00$
C) $x = 6.45$
D) $x = - 7.45; x = 6.45$
E) none of these

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